Question

A student collects 350 mL of a vapor at a temperature of 67°C. The atmospheric pressure at the time of collection is 0.900 atm. The student determines that the mass of the vapor collected is 0.79 g. Which of the following is most likely to be the chemical analyzed?
acetone
ethanol
ethyl acetate
pentane.

Answer 1

Answer:

The answer to your question is: Pentane (0.011 moles)

Explanation:

Data

V = 350 ml = 0.35 l

T = 67°C = 340 °K

P = 0.9 atm

mass = 0.79 g

R = 0.082 atm L/mol°K

Formula

                       PV = nRT

                       n = PV / RT

                       n = (0.9)(0.35) / (0.082)(340)

                       n = 0.315 / 27.88

                      n = 0.0112

Now

MW acetone = 58g

MW ethanol = 46g

MW ethyl acetate = 88 g

MW pentane = 72 g

For acetone

                 58 g ------------ 1 mole

                 0.79 g --------   x

                 x = 0.014 moles

For ethanol

                46g --------------- 1 mole

                0.79g  ------------ x

               x = 0.17 moles

For ethyl acetate

                88 g ------------- 1 mole

                0.79 g -----------  x

                x = 0.0089 moles

For pentane

              72 g -------------- 1 mole

              0.79 g ------------  x

             x = 0.011 moles

The substance is pentane